On Sun, Dec 16, 2001 at 03:55:10PM +1100, Damian Conway wrote:
[...]
And, just for laughs:
$ref = [1,2];
@ary[$ref] = foo(); # probably a syntax error
Ok, as far as I can recall, Larry hinted that arrays and references to
arrays would be interchangable in many
Aaron Sherman [EMAIL PROTECTED] writes:
On Sun, Dec 16, 2001 at 03:55:10PM +1100, Damian Conway wrote:
[...]
And, just for laughs:
$ref = [1,2];
@ary[$ref] = foo(); # probably a syntax error
Ok, as far as I can recall, Larry hinted that arrays and references to
$val = (foo())[0];
List?
Scalar, obviously.
How do you figure that? (Not a criticism: I'd really like to understand your
thought process here so I can assess the relative DWIMity of the two
alternatives).
With a possible runtime error if foo doesn't return
Aaron Sherman wrote:
$ref = [1,2];
@ary[$ref] = foo(); # probably a syntax error
Ok, as far as I can recall, Larry hinted that arrays and references to
arrays would be interchangable in many contexts in P6. In this case, I
can't see any reason that subscripting would *want*
Damian Conway [EMAIL PROTECTED] writes:
$val = (foo())[0];
List?
Scalar, obviously.
How do you figure that? (Not a criticism: I'd really like to understand your
thought process here so I can assess the relative DWIMity of the two
alternatives).
I figure
Piers posed the following puzzles:
@ary[0] = foo() # scalar
Yes.
@ary[1,2] = foo() # list context
Yes.
@bar = 1;
@ary[@bar] = foo() # ? probably list or maybe scalar...
List. With an explicit array as index, it's definitely a (one-element) slice.
@bar = (1,2);
In the following code fragment, what context is foo() in?
@ary[0] = foo()
the following code
@ary= foo()
obviously evaluates @foo in a list context, but in the first I'm no
longer sure.
--
Piers
It is a truth universally acknowledged that a language in
possession of a
Thus it was written in the epistle of Piers Cawley,
In the following code fragment, what context is foo() in?
@ary[0] = foo()
Scalar, I would think.
Just my guess,
Ted
--
Ted Ashton ([EMAIL PROTECTED]) | From the Tom Swifty collection:
Southern Adventist University| Multiplication
On Thu, Dec 13, 2001 at 12:12:14PM -0500, Ted Ashton wrote:
Thus it was written in the epistle of Piers Cawley,
In the following code fragment, what context is foo() in?
@ary[0] = foo()
Scalar, I would think.
I assume that the following would make the assignment a slice
and
Piers Cawley:
# In the following code fragment, what context is foo() in?
#
# @ary[0] = foo()
The short answer is scalar context. The long answer is below. Note
that the long answer is only the way I think of it. You may think
differently.
I like to think of it as 'one context'.
In the following code fragment, what context is foo() in?
@ary[0] = foo()
Scalar context. @ary[0] is a single element of @ary.
To call foo() in list context use any of the following:
(@ary[0]) = foo(); # Assign @ary[0] the first element returned
On Fri, Dec 14, 2001 at 06:39:02AM +1100, Damian Conway wrote:
In the following code fragment, what context is foo() in?
@ary[0] = foo()
Scalar context. @ary[0] is a single element of @ary.
To call foo() in list context use any of the following:
(@ary[0]) =
@ary[0] =()= foo(); #
Hm, thats a change from perl5. In perl5 that would assign the number of
elements returned from foo(). Is there a good reason for this change ?
Firstly, Larry may have to rule on which behaviour actually *is* invoked
Brent Dax [EMAIL PROTECTED] writes:
Piers Cawley:
# In the following code fragment, what context is foo() in?
#
# @ary[0] = foo()
The short answer is scalar context. The long answer is below. Note
that the long answer is only the way I think of it. You may think
differently.
I
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